更新时间:2022-02-12 08:35:55
哈斯克尔学校, LYAH , Quora 以及更多描述单子的列表.
Wadler, School of Haskell, LYAH, HaskellWiki, Quora and many more describe the list monad.
比较:
(=<<) :: Monad m => (a -> m b) -> m a -> m b
用于具有concatMap :: (a -> [b]) -> [a] -> [b]
用于m = []
.(=<<) :: Monad m => (a -> m b) -> m a -> m b
for lists withconcatMap :: (a -> [b]) -> [a] -> [b]
for m = []
.常规的(>>=)
绑定运算符的参数已翻转,但否则只是一个后缀concatMap
.
The regular (>>=)
bind operator has the arguments flipped, but is otherwise just an infix concatMap
.
或者很简单地,我的困惑似乎是由于不了解该语句的实际工作原理而产生的:
Or quite simply my confusion seems to stem from not understanding how this statement actually works:
(>>|) xs f = [ y | x <- xs, y <- f x ]
由于列表推导与列表的Monad实例等效,因此此定义是一种作弊行为.基本上,您说的是某种东西,就像它是Monad一样,是Monadd,所以剩下两个问题:了解列表推导和仍然了解Monad.
Since list comprehensions are equivalent to the Monad instance for lists, this definition is kind of cheating. You're basically saying that something is a Monadd in the way that it's a Monad, so you're left with two problems: Understanding list comprehensions, and still understanding Monad.
可以理解列表的理解力,以便更好地理解:
List comprehensions can be de-sugared for a better understanding:
对于您而言,该语句可以用多种其他方式编写:
In your case, the statement could be written in a number of other ways:
使用符号:
Using do-notation:
(>>|) xs f = do x <- xs
y <- f x
return y
建议使用(>>=)
运算符:
De-sugared into using the (>>=)
operator:
(>>|) xs f = xs >>= \x ->
f x >>= \y ->
return y
这可以缩短(每行重写一次):
This can be shortened (one rewrite per line):
(>>|) xs f = xs >>= \x -> f x >>= \y -> return y -- eta-reduction
≡ (>>|) xs f = xs >>= \x -> f x >>= return -- monad identity
≡ (>>|) xs f = xs >>= \x -> f x -- eta-reduction
≡ (>>|) xs f = xs >>= f -- prefix operator
≡ (>>|) xs f = (>>=) xs f -- point-free
≡ (>>|) = (>>=)
因此,通过使用列表推导,您尚未真正声明一个新的定义,而是仅依赖现有的定义.如果需要,您可以定义instance Monadd []
而不依赖现有的Monad实例或列表理解:
So from using list comprehensions, you haven't really declared a new definition, you're just relying on the existing one. If you wanted, you could instead define your instance Monadd []
without relying on existing Monad instances or list comprehensions:
使用concatMap
:
instance Monadd [] where
(>>|) xs f = concatMap f xs
进一步说明:
Spelling that out a little more:
instance Monadd [] where
(>>|) xs f = concat (map f xs)
进一步说明:
Spelling that out even more:
instance Monadd [] where
(>>|) [] f = []
(>>|) (x:xs) f = let ys = f x in ys ++ ((>>|) xs f)
Monadd类型类应具有与return
类似的内容.我不确定为什么会丢失它.
The Monadd type class should have something similar to return
. I'm not sure why it's missing.